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4 years ago

A railroad track rises 30 feet for every 400 feet of track. What is the measure of the angle of elevation of the track?

Mathematics

1 answer:

gtnhenbr [62]4 years ago

3 0

Answer:

Step-by-step explanation:

Rising 30 feet for every 400 feet of track is not a very large angle of elevation. We set this up as a right triangle to use right triangle trig to solve for the missing angle. The height is 30 and the hypotenuse is 400. The height is the side opposite the angle in question, and the hypotenuse is the hypotenuse! This is the sin ratio. Setting up the ratio using our info:

$sin\theta=\frac{30}{400}$

When you are looking for a missing angle measure, as we are here, use the 2nd-->sin buttons to find the inverse of sin, which looks like this:

$sin^{-1}($

Enter in the fraction 30/400 after the parenthesis and then hit enter to get the angle measure of 4.3 degrees.

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A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

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Larger number = (sum + difference)/2
Smaller number = (sum - difference)/2

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